Let $H$ and $O$ be the orthocenter and circumcenter of $A B C$, respectively: it is well-known (and not difficult to check) that $\measuredangle$ B A $H=\measuredangle$ C A O$. However, note that \measuredangle B A H=\measuredangle B A D=\measuredangle C A F$, so $\measuredangle$ C A $F=\measuredangle$ C A O$, that is,$O$lies on median$A F$, and since$A B<A C$, it follows that$F=O$. Therefore, \measuredangle B A C=90^{\circ}$. Now, we compute $\measuredangle$ A C $B=\measuredangle$ B A $D=\frac{2}{6} \measuredangle$ B A $C=30^{\circ}\$.$